1. Statics: Trans. and Tors.

2. Force Equilibrium: 1.Draw Picture 2.Calculate weights 3.Express/calculate components 4.Set up a = 0 equation for x and another for the y direction Torque Equilibrium: 1.Pick a Pivot Point (at location of unknown force) 2.Express all torques: 3.+rF +rF+rF… = 0 + is CW, - is ACW r is distance from pivot Do Maths

3. Force Equilibrium: 1.Draw Picture 2.Calculate weights 3.Express/calculate components 4.Set up a = 0 equation for x and another for the y direction Torque Equilibrium: 1.Pick a Pivot Point (at location of unknown force) 2.Express all torques: 3.+rF +rF+rF… = 0 + is CW, - is ACW r is distance from pivot Do Maths Example 1: Find F1 and F2 64.0 kg F1F1 F2F2 The beam is 6.0 m long, 45 kg and uniform. The person is standing 0.50 m from the right side, and F 2 is 4.0 m from the left side -125.08 N (down), 1194.4 N (up)

4. Torque: (0 m)F1 – (4 m)F2 + (3 m)(45*9.81 N) + (5.5 m)(64*9.81 N) = 0 F2 = 1194.4 N X: - nada Y: F1 + F2 - 45*9.81 - 64*9.81 = 0 F1 + 1194.4 -45*9.81 - 64*9.81 = 0 F1 = -125.1 N Example 1: Find F1 and F2 64.0 kg F1F1 F2F2 The beam is 6.0 m long, 45 kg and uniform. The person is standing 0.50 m from the right side, and F 2 is 4.0 m from the left side

5. Force Equilibrium: 1.Draw Picture 2.Calculate weights 3.Express/calculate components 4.Set up a = 0 equation for x and another for the y direction Torque Equilibrium: 1.Pick a Pivot Point (at location of unknown force) 2.Express all torques: 3.+rF +rF+rF… = 0 + is CW, - is ACW r is distance from pivot Do Maths 350 kg 85 kg Example 2: Find The tension in the cable The Force of the wall in the x dir The Force of the wall in the y dir 58 o 12.0 m 1.0 m 5.0 m 2655.3 N, 1407.1 N right, 2105.5 N up

6. Torque: (0)Wy + (5 m)(85*9.81 N) + (6 m)(350*9.81 N) – (11 m)Tsin(58) = 0 T = 2655.32 N X: Wx – Tcos(58) Wx = 1407.11 N Y: Wy – 85*9.81 – 350*9.81 + Tsin(58) = 0 Wy = 2015.5 N 350 kg 85 kg 58 o 12.0 m 1.0 m 5.0 m 2655.3 N, 1407.1 N right, 2015.5 N up Example 2: Find The tension in the cable The Force of the wall in the x dir The Force of the wall in the y dir

7. Whiteboards: Trans and Tors 1 2 3 4 TOC

8. T 2 = 801 N, T 1 = 465 N 52 kg 77 kg T1T1 T2T2 Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end. +4586.4 Nm + 9809.8 Nm = (18.0 m)(T 2 ), T 2 = 799.8 N T 1 + 799.8 N -509.6 N - 754.6 N = 0, T 1 = 464.4 N

9. T = 181 N, Fy = 309 N up W 35 kg 15 kg T 7.0m 16.0m 20.0m (Fulcrum) Find the tension in the cable and the upward force from the fulcrum. (hint – torque about the fulcrum – you will need to figure out the distances)

10. T = 375 N, Wx = 325 N (right), Wy = -20.9 N (down) The beam is uniform and 4.00 m long, the cable is attached 2.30 m from the left side at a 30.0 o angle with the beam Find the tension in the cable and the force exerted by the wall in the x and y direction 12.0 kg 5.00 kg

11. T = 274.8 N, Wx = 274.8 N right, Wy = 235.4 N up The 10.0 kg beam is uniform and 8.00 m long and makes an angle of 35.0 o with the wall, the cable is attached 3.80 m from the bottom end. Find the tension in the cable and the force exerted by the wall in the x and y direction (Careful what angle you use for the tension) 14.0 kg